# How to get ripple voltage of dc to dc converter?

#### Saviour Muscat

Joined Sep 19, 2014
147
Dear members

Hope this post finds you in good health in these festive days.
I am doing a question about dc to dc converter given by the lecturer.
The question is as follows
Design step down DC converter to convert 12V dc to 5V dc at the load. Current ripple should be less than 0.27A and switched at frequency of 10kHz. The voltage ripple output must be less than 10mV and the load is a 5ohm resistor.

I did some calculations and assumed inductor about 10mH and found the Ripple current about 0.0489A.
I am not sure if I will get the value(ripple voltage) from simulation or from calculation(kindly indicate the equation) or indicate the best way please.
Please refer to the drawn circuit.

Thanks
SM

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#### Jony130

Joined Feb 17, 2009
5,163
Why didn't you use an output capacitor?

• Saviour Muscat

#### Saviour Muscat

Joined Sep 19, 2014
147
Why didn't you use an output capacitor?
Because the notes we did in class are without a capacitor.
The best way is it by calculation(I don't know which equation could I use) or by simulation?

#### MrAl

Joined Jun 17, 2014
7,588
Dear members

Hope this post finds you in good health in these festive days.
I am doing a question about dc to dc converter given by the lecturer.
The question is as follows
Design step down DC converter to convert 12V dc to 5V dc at the load. Current ripple should be less than 0.27A and switched at frequency of 10kHz. The voltage ripple output must be less than 10mV and the load is a 5ohm resistor.

I did some calculations and assumed inductor about 10mH and found the Ripple current about 0.0489A.
I am not sure if I will get the value(ripple voltage) from simulation or from calculation(kindly indicate the equation) or indicate the best way please.
Please refer to the drawn circuit.

Thanks
SM
Hi,

Normally we do see an output capacitor. However, if this is the way the instructor wants you to do it then you have to follow that lead. The output may be somewhat normal anyway except the output voltage ripple is always lower with an output capacitor.

What kind of analysis have you done in the past?
For example, differential equations, ODE's, Laplace Transforms.
Also, do you know how to calculate the theoretical duty cycle knowing the input and output voltages?

And how did you calculate the ripple current?

This is a fairly simple circuit so it should not be hard to come up with something.

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• Saviour Muscat

#### BobTPH

Joined Jun 5, 2013
2,388
Given a load resistance and a current ripple, can you come up with an expression for the voltage ripple?

Bob

• Saviour Muscat

#### Saviour Muscat

Joined Sep 19, 2014
147
Thank you all for your help
Below is there some working that I did

Vout=DVin
5=D12
D=5/12=41.7%

Taw=L/R=10m/5=2ms
T=1/f=1/10k=0.1ms
Irpp=Vin/R *(1/(1-e^-T/Taw))*((1+e^-T/Taw)-(e^-DT/Taw)+(e^-(1-D)T/Taw))
Irpp=0.04877A

#### MrAl

Joined Jun 17, 2014
7,588
Thank you all for your help
Below is there some working that I did

Vout=DVin
5=D12
D=5/12=41.7%

Taw=L/R=10m/5=2ms
T=1/f=1/10k=0.1ms
Irpp=Vin/R *(1/(1-e^-T/Taw))*((1+e^-T/Taw)-(e^-DT/Taw)+(e^-(1-D)T/Taw))
Irpp=0.04877A
Your result might be right but not reproducible from your expression of Irpp.
Irpp must be written out wrong.

#### Saviour Muscat

Joined Sep 19, 2014
147
Again thank you!
Sorry I made a mistake in the formula Irpp
I forgot Two brackets

#### crutschow

Joined Mar 14, 2008
24,938
Design step down DC converter to convert 12V dc to 5V dc at the load. Current ripple should be less than 0.27A and switched at frequency of 10kHz. The voltage ripple output must be less than 10mV and the load is a 5ohm resistor.
That's not possible without an output filter capacitor.

#### BobTPH

Joined Jun 5, 2013
2,388
Sure it is, but the inductor might be impractically large.

Bob

#### MrAl

Joined Jun 17, 2014
7,588
That's not possible without an output filter capacitor.
Hi,

Why not?

#### MrAl

Joined Jun 17, 2014
7,588
Again thank you!
Sorry I made a mistake in the formula Irpp
I forgot Two brackets
Hi,

Ok but i get a different result than you did then.
My result matches two other techniques i use:
1. Regular simulator like LT Spice.
2. "Analytical Simulation" using the analytical time equations for the two modes of switch operation (on and off).
(Actually there was one more technique purely analytical and not time averaged per se' but the time equations are the same as #2 above).
There is also another technique which is derived straightforward from the 'on' time equation which comes within 0.5 percent of the more precise result. It's an estimate.
3. AC averaged solutions. Ok that's three distinctly different.

The first two enumerated techniques above #1 and #2 provide the same result i got with YOUR expression which is just under 0.030 amps (i wont give the exact result yet until you had a chance to try it again). The third #3 comes very very close.

If you know Ohm's Law then you know how to calculate the output voltage ripple also. Can you see how to do this?

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#### Saviour Muscat

Joined Sep 19, 2014
147
Thanks a million to all,
These days I downloaded a spice simulator as suggested by MrAL and I tried out the circuit on it.
I ran the simulation several times varying the inductance of the inductor so to get about 10mV ripple and the most appropriate value that I found was about 120mH ass seen from the screenshots. Then I calculated the corresponding value of the Irpp which resulted about 2.43mA. Please correct me if did wrong!
Thanks and best wishes,
SM

#### MrAl

Joined Jun 17, 2014
7,588
Thanks a million to all,
These days I downloaded a spice simulator as suggested by MrAL and I tried out the circuit on it.
I ran the simulation several times varying the inductance of the inductor so to get about 10mV ripple and the most appropriate value that I found was about 120mH ass seen from the screenshots. Then I calculated the corresponding value of the Irpp which resulted about 2.43mA. Please correct me if did wrong!
Thanks and best wishes,
SM
Hi,

Well, if you have 2.43ma peak to peak current ripple then how could you have 10mv peak to peak voltage ripple with a 5 Ohm output resistor load when the only current through the resistor is the current in the inductor?

The simulator is good for checking your results, but you should also do the calculations by hand to get a feel for how to do these problems. There will eventually (in the future) be results you will need that you cant get easily with a simulator.

[LATER]
Also note that sometimes the values from a simulation may be hard to read to get good accuracy or precision. Doing it by hand you can get easily 10 digit precision in many cases. This is helpful when comparing for example the analytical time equation solutions to the the AC averaged solutions ... there is a tiny difference which would be hard or impossible to notice with a simulation.

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#### Saviour Muscat

Joined Sep 19, 2014
147
Thank you for your kind help and attention

I need to design snubber circuits during turn off and turn on, could you advise books or articles to read with worked examples so to be able to design the snubber circuit, please?

Again thanks and I wish you all the best in the new year!
SM

#### MrAl

Joined Jun 17, 2014
7,588
Hi,

I did a google of this:
"design of snubber circuit"
and all kinds of solutions came up included a pdf.
If you find one you would like to use we can talk about it here if you like. Post the schematic.

#### Saviour Muscat

Joined Sep 19, 2014
147
Hello members,

I redesign the DC to DC converter using a smoothing capacitor across the load because as Bob said the inductor will be very large.
I used these three equations suggested from Jony notes.

1. Vout=DVin D=5/12

1. 2. L=((Vin-Vout)*ton)/Irpp, where Irpp=0.29A, T=1/10000., L=1.00574mH

2. 3. C=((Vin-Vout)*(ton)^2)/(2*L*Vrpp), where Vrpp=10mV and L=1.00574mH, C=604.2uF

With these results, I simulated in Spice simulator and worked fine

The problem that I encountered How to design the RC snubber across Collector emitter pins of the transistor
I did a tentative circuit as suggested by MrAL , and by trial and error the current spikes flowing from collector were minimized but I am not convinced If I did well, please could someone help me to calculate the components and show me if it is possible a numerical example because I am lost!

Thanks
SM

#### MrAl

Joined Jun 17, 2014
7,588
Hello members,

I redesign the DC to DC converter using a smoothing capacitor across the load because as Bob said the inductor will be very large.
I used these three equations suggested from Jony notes.

1. Vout=DVin D=5/12

1. 2. L=((Vin-Vout)*ton)/Irpp, where Irpp=0.29A, T=1/10000., L=1.00574mH

2. 3. C=((Vin-Vout)*(ton)^2)/(2*L*Vrpp), where Vrpp=10mV and L=1.00574mH, C=604.2uF

With these results, I simulated in Spice simulator and worked fine

The problem that I encountered How to design the RC snubber across Collector emitter pins of the transistor
I did a tentative circuit as suggested by MrAL , and by trial and error the current spikes flowing from collector were minimized but I am not convinced If I did well, please could someone help me to calculate the components and show me if it is possible a numerical example because I am lost!

Thanks
SM
Hi,

I thought this was school work and you had to use just an inductor with no capacitor?

Yes sure a capacitor will help the output voltage ripple that's why most converters have them.

I'll take a look at all this new material.